## Review of Thermodynamics

You should look at the following pages and understand everything about the page. The pages are the freshman chemistry material on thermodynamics.

General Chemistry Theromodynamics Chapter

## Review of Newton's Law of Mechanics

**1 ^{st} law:** A body remains at rest or in uniform motion unless acted upon by a force

**2**

^{nd}law:*F = ma*A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force

**3**

^{rd}law:*F*If two bodies exert forces on each other, these forces are equal in magnitutde and opposite direction

_{1}= -F_{2}

### Energy

**Total Energy**is composed of kinetic (

*T*) and potential (

*V*) energies

$$T=\frac{1}{2}m{v}^{2}$$ | 1 |

$$V=V\left(x\right)$$ | 2 |

**Force**, appeared in Newton's 2^{nd} law, can be defined as

$$F=-\frac{\partial V}{\partial x}$$ | 3 |

*V*is the potential energy and

*x*is the coordinate. One of chemical applications includes molecular vibration. In molecular vibration,

*x*is given as the nuclear coordinates and

*V*is the so-called

*potential energy surface*with

*3N - 6*(

*N*is the number of atoms) dimensions, describing how energy change with respect to coordinates, such as bond distance, bond angle and dihedral angle, among others.

The following picture illustrates what the force is. You can think about
the *x* axis as a reaction coordinate for which
reaction proceeds from left to right. The energy change along the reaction
coordinate is represented by *V*.

**Figure 1. Relationship between potential energy and force.**

As is seen in the figure, the tangent line of *V* at a point
*x _{i}* is shown as positive slope. Therefore, the force
exerted is negative of the tangent line. The force points to the
negative direction. Also, steeper the tangent line, larger the force is.

*empirically derived*potential energy forms available. These potential energy forms are for molecular motion or molecular vibration, thus deal with how energy changes with respect to changes in atomic positions.

Most fundamental of all is potential energy curve of diatomic molecules. Since there is only one coordinate, stretching coordinate, one can represent potential energy curve as a function of the bond stretching coordinate. Here one prominent example of diatomic molecular potential called Morse potential is given.

**Morse Potential**:
$$V\left(R\right)={D}_{e}{\left(1-{e}^{-a\left(R-{R}_{e}\right)}\right)}^{2}$$
where *D _{e}* is the dissociation energy of a molecule,

*R*is the equilibrium bond distance, and

_{e}*a*is defined as $$a={\left(\frac{{k}_{e}}{2{D}_{e}}\right)}^{1/2}$$ where

*k*is the force constant, typically given in units such as mdyn/Å or dyn/cm, where 1 dyn = 10 μN = 10

_{e}^{-5}kg m / s

^{2}. The force constant is related to the vibrational frequency via $$\nu =\frac{1}{2\pi}{\left(\frac{{k}_{e}}{\mu}\right)}^{1/2}$$ where μ is a reduced mass of the molecule, which is $$\mu =\frac{{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}$$ Diatomic potential energy curve looks like the following, including the important features that are in the Morse potential.

The lowest point of the potential is given by *D _{e}*,
representing the

*dissociation energy*of the molecule, and it occurs at

*equilibrium distance, R*or

_{e}*equilibrium bond distance*. Notice that the interaction energy is a

*negative*value, indicating stability relative to the separated atoms. When the bond is stretched to the point where the interaction between two atoms are small, it is the right-hand region where the

*V(R)*tapers down to value of 0. Zero because there is no interaction.

On the opposite side of *R _{e}*, that is to shrink the bond,
compared to

*R*, the intection energy goes up, eventually becomes positive number, meaning that the interacton energy is repulsive.

_{e}

You can see the lines *ν _{0}* and

*ν*. These are the quantized vibrational energy levels of diatomic molecules. In order to make a transition from the lowest energy state,

_{1}*ν*to an excited state,

_{0}*ν*, an infrared photon is aborbed by the molecule, which is registered in IR (or Raman) spectrometer.

_{1}

The equation that relates frequency and force constant is actually an
approximation equation. As long as you only look at fundamental transition,
which is ν_{1} ← *ν _{0}
*, the approximation is quite good.

For example, the force constant of H

_{2}molecule is 5.756 mdyn/Å, we can calculate vibrational frequency using above equations to be 4403 cm

^{-1}, which is in agreement with the Raman spectra.

Brief breakdown is as follows:

5.756 mdyn/Å = 5.756 x 10^{2} kg/s^{2}

μ = (1.008 amu)^{2}/(1.008 + 1.008 amu) = 0.504 amu

Using a conversion factor of 1.66053892 x 10^{-27} kg = 1 amu,
0.504 amu = 0.83691 x 10^{-27} kg.

ν = (1/2π)(5.756 x 10^{2} kg/s^{2}/0.83691 x
10^{-27} kg)^{1/2} = 1.3199 x 10^{14} /s or Hz

Using a conversion factor of 1 Hz = 3.33565 x 10^{-11}
cm^{-1}, one obtains 4403 cm^{-1}

## Review of Electricity and Magnetism

We know that atom is made of nucleus and electrons for which they possess electrical charges, plus and minus, respectively. Absorption and emission of light, which is electromagnetic wave, is also characteristics of atoms. Therefore, understanding electricity and magnetism is quite important.

### Interaction of Charged Particles

Let's start by looking at interaction of charged particles. The potential energy between charged particles*i*and

*j*are denoted as

*V*, and is given by

_{ij}*Coulomb's Law*.

$${V}_{ij}=\frac{1}{k}\frac{{q}_{i}{q}_{j}}{{r}_{ij}}$$ | 4 |

*q*and

_{i}*q*are the electrical charges of particle

_{j}*i*and

*j*, respectively, and

*k*is a constant (

*k*= 4

*πε*with

_{0}*ε*is called

_{0}*vaccuum permitivity*). If

*q*and

_{i}*q*have the same charges, then the

_{j}*V*is positive, indicating that the interaction is

*repulsive*. If

*q*and

_{i}*q*have the opposite charges, the interaction

_{j}*V*becomes

*attractive*.

### Magnetism

Magnet has two poles, north and south. A directional compus points to magnetic north pole of earth. In analogy with repulsion and attraction in electrostatics, like magnetic poles repel, and different poles attract.

In materials, there are several magnetism occurs.

### Magnetic field produced by current

Classic experiment by Orsted in 1820 shows electrical current can induce magnetic field on the plane perpendicular to the wire, as shown below. Directional compasses are placed on the bench where the electrical wire is placed perpendicular direction. When a current flows through the wire, compasses point in circularly pointed.

**Figure 2. Magnetic field produced by electrical current**

When a current is off, all compasses point to north, as usual. When, however,
a current is turned on, the compasses point in circular pattern, represented
by a so-called, *right hand rule*, as shown in Figure 3.

**Figure 3. Result of Oersted experiment**

In the above example, the right panel of Figure 3, the current is going away from the screen to the back-side of the monitor you're looking at. One can calculate magnetic field around the wire, as

$$B=\frac{{\mu}_{0}I}{2\pi r}$$ | 5 |

*μ*is vacuum permeability,

_{0}*I*is current, and

*r*is the distance from the wire.

### Force between Two Wires

If two wires are parallel to each other and the current
flows through both wires, you observe that there appears either
repulsion or attraction on the wires. Using Eqn. 5, the distance *
r* is set to the distance between wires, the force exerted on one
wire is given by

$$F=IBl$$ | 6 |

*B*. The force that is feld by a segment of wire 1 carrying a current

*I*

_{1}$${F}_{1}={I}_{1}l{B}_{2}$$ | 7 |

*B*, we get

_{2}$$F={\mu}_{0}l\frac{{I}_{1}{I}_{2}}{2\pi r}$$ | 8 |

*r*is the distance between two wire of the length

*l*. If the current in the two wires flow in the same direction, the force between the wires are attractive in nature, as shown in Figure 4. When the current is reversed in one wire, the force between the two wires are repulsive.

**Figure 4. Force between two wires with current.**

The small line segment *l* for magnetic field and its direction
are also shown in the inset in the figure. From the orientation of
these magnetic fields genearted by two wires
can attest to the repulsive and attractive natures of two cases.

We can put Eqn. 5 into integral form as,

$$\oint B\cdot dl={\mu}_{0}I$$ | 9 |

### Faraday's Law of Electrical Induction

Since a current induces a magnetic field, it is possible that the opposite can happen. About 10 years after Oersted's experiment, Henry and Faraday independently examined the electrical induction effect by magnet. The following figure shouws how an experiment can be carried out.**Figure 5. Current produced by moving magnet.**

An electrical wire is wound up in a coil and is connected to an ampmeter,
which registers the current induced by motion of a magnet. When a magnet goes
through the loop of wire, the current *I* is induced.

The direction to which the current flow is given by *Lenz Law*,
which states that a current will flow due to the induced magnetic field
opposing the magnetic field. Use a right-hand rule, when N side of the magnet
approaches the loop, as shown in Figure 5b, the induced magnetic field
closer to the N side of the magnet becomes N. Thus, the magnetic field
develped in the loop is counterclockwise by pointing your thumb to you,
the curling of other fingers tell you the couterclockwise motion.

When the magnet is pulling away from the loop as shown in Figure 5c, then
the induced magnetic field point toward the rear of the monitor you're
looking at. It means that the side closest to the magnet is S. Using the
right-hand rule, tell you that the current induced is clockwise.
If we define *magnetic flux* to be the sum of magnetic line passing
through an area, *A*, is

$${\Phi}_{B}=\int B\cdot dA$$ | 10 |

*electromotive force*is related to

*electric field*

$$E=-\frac{d{\Phi}_{B}}{dt}$$ | 11 |

*dl*for electric field,

$${\Phi}_{E}=\oint E\cdot dl$$ | 12 |

$$\oint E\cdot dl=-\frac{d{\Phi}_{B}}{dt}$$ | 13 |

### Displacement Current

A magnetic field is produced by not only by an electrical current. When a capacitor is placed in a circuit, accumulation of charges on the parallel plates should also give rise to magnetic field. Since accumulation of charge on the capacitor plates changes the electric fields,

The charge on a capacitor is given by

$$Q=CV$$ | 14 |

*C*is capacitance and

*V*is the potential difference between the two plates and

*V = Ed*with

*E*is the electric field and

*d*is the distance between the plates. The capacitance is given by

$$C={\u03f5}_{0}\frac{A}{d}$$ | 15 |

*A*is the area of each plate of the capacitor. Therefore,

*Q*is given by

$$Q={\u03f5}_{0}AE$$ | 16 |

$$\frac{dQ}{dt}={\u03f5}_{0}A\frac{dE}{dt}={\u03f5}_{0}\frac{d{\Phi}_{E}}{dt}$$ | 17 |

*Φ*is the electric flux defined by

_{E}*Φ*.

_{E}= EA

Then, we can write Maxwell equation for Ampere's law to be

$$\oint B\cdot dl={\mu}_{0}\left(I+{\u03f5}_{0}\frac{d{\Phi}_{E}}{dt}\right)$$ | 18 |

### Gauss's Law

The electric flux calulated within a closed area on a surface is the net charge,*Q*,

$$\oint E\cdot dA=\frac{Q}{{\u03f5}_{0}}$$ | 19 |

$$\oint B\cdot dA=0$$ | 20 |

*A*is zero, which means that there is no magnetic equivalent of single electric charges,

*monopoles*.

### Simple Harmonic Motion

The force associated with harmonic motion is given by$$F=-kx$$ | 21 |

*F=ma*, the differential equation that represents harmonic motion is

$$m\frac{{d}^{2}x}{d{t}^{2}}=-kx$$ | 22 |

$$\frac{{d}^{2}x}{d{t}^{2}}+\frac{k}{m}x=0$$ | 23 |

$$x=Acos\left(\omega t+\varphi \right)$$ | 24 |

*A*is the amplitude or the maximum displacement of mass

*m*. The arguments of cosine function is

*ω*is the frequency of the sinusoidal motion (or peak to peak frequency), φ is a constant phase angle to shift the wave.

### Traveling Waves

Traveling wave can be represented by sinosoidal functions as we have seen above for harmonic oscillator solution. In the traveling wave case, the wave*Φ*(

*x,t*) at certain coordinate

*x*is moving with a speed

*v*, as shown in Figure 6.

**Figure 6. Traveling wave.**

At some time *t* = 0, let's say that the wave is represented by the
black curve, but when at some time has passed, the wave has shifted due to
its velocity, shown as red curve in the figure.
In this case, *Φ*(*x,t*) is represented by

$$\Phi \left(x,t\right)={\Phi}_{0}sin\left[\frac{2\pi}{\lambda}\left(x-vt\right)\right]$$ | 25 |

*Φ*is the maximum displacement of the wave amplitude, λ is the wave length and

_{0}*v*is the velocity of the wave.

### Propagation of Electromagnetic Waves

We use Eqns. 13 and 18 of Maxwell's equations to explain how electromagnetic wave propagates. Consider a small rectangular box (purple colored box) labeled with*dx*and

*dy*in the electric field (wave in blue) in Figure 7.

**Figure 7. Propagation of Electromagnetic Wave.**Electric=Blue Magnetic=Red

Since the magnetic field is increasing in time in the process in the
purple rectangle, the induced magnetic field in the reactangle decreases
according to Lenz law, then the electric field
Let us examine the propagation using Eqns. 13 and 18.
The idea is to examine what's happening with the purple areas in Figure 7.
In Figure 7a, the induced magnetic field in the purple loop should decrease
in time with the wave moving toward +*x* direction. Then, the electric
field should point opposite to this change. Therefore, the electric
field is greater on the right side of the loop, a counterclockwise
circulation (or current without electron, if you want!). The circulation
produced is produced to oppose the change in *Φ _{B}*.

$${E}_{y}={E}_{0}sin\left(kx-\omega t\right)$$ | c1 |

$${B}_{z}={B}_{0}sin\left(kx-\omega t\right)$$ | c2 |

*k*,

*ω*and velocity

*v*are defined respectively as,

$$k=\frac{2\pi}{\lambda}$$ | c3 |

$$\omega =2\pi f$$ | c4 |

$$\omega \lambda =v$$ | c5 |

*λ*is wave length and

*f*is frequency. Eqns c1 - c5 are assumed to be the solution to the Maxwell equations, Eqn. 13 and 18, and are

$$\oint E\cdot dl=-\frac{d{\Phi}_{B}}{dt}$$ | 13 |

$$\oint B\cdot dl={\mu}_{0}\left(I+{\u03f5}_{0}\frac{d{\Phi}_{E}}{dt}\right)$$ | 18 |

$$\oint E\cdot dl=\left(E+dE\right)\Delta y-E\Delta y=dE\Delta y$$ | c6 |

*dx*is perpendicular to the electric field,

*E*, the

*dx*component is zero, and only contributions come from

*Δy*. The right-hand side of Eqn. 13 is

$$-\frac{d{\Phi}_{B}}{dt}=-\frac{dB}{dt}dx\Delta y$$ | c7 |

*Phi;*(Eqn. 10). Therefore, we have

_{B}$$dE\Delta y=-\frac{dB}{dt}dx\Delta y$$ | c8 |

$$\frac{dE}{dx}=-\frac{dB}{dt}$$ | c9 |

There is no electrical current, therefore the current term on the right-hand side of Eqn. 18 is neglected, in below. The arguments are similar to the one for Eqn. 13 and one can derive,

$$\oint B\cdot dl={\mu}_{0}{\u03f5}_{0}\frac{d{\Phi}_{E}}{dt}$$ | c10 |

$$\oint B\cdot dl=B\Delta z-\left(B+dB\right)\Delta z=-dB\Delta z$$ | c11 |

$${\mu}_{0}{\u03f5}_{0}\frac{d{\Phi}_{E}}{dt}={\mu}_{0}{\u03f5}_{0}\frac{dE}{dt}dx\Delta z$$ | c12 |

$$\frac{dB}{dx}=-{\mu}_{0}{\u03f5}_{0}\frac{dE}{dt}$$ | c13 |

*E*and

_{0}*B*in Eqns. c1 and c2. According to Eqn. c9, taking the (partial) derivative of

_{0}*E*with respect to

*x*and taking the partial derivative of

*B*with respect to

*t*are equal to each other,

*i.e.*,

$$k{E}_{0}cos\left(kx-\omega t\right)=\omega {B}_{0}cos\left(kx-\omega t\right)$$ | c14 |

*E*and

_{0}*B*is,

_{0}$$\frac{{E}_{0}}{{B}_{0}}=\frac{\omega}{k}=v$$ | c15 |

*B*with respect to

*x*on the left-hand side and taking the partial derivative of

*E*with respect to time give,

$$k{B}_{0}cos\left(kx-\omega t\right)=\omega {\mu}_{0}{\u03f5}_{0}{E}_{0}cos\left(kx-\omega t\right)$$ | c16 |

$$\frac{{B}_{0}}{{E}_{0}}=\frac{\omega {\mu}_{0}{\u03f5}_{0}}{k}={\mu}_{0}{\u03f5}_{0}v$$ | c17 |

$$\frac{{B}_{0}}{{E}_{0}}=\frac{1}{v}={\mu}_{0}{\u03f5}_{0}v$$ | c18 |

$$v=\frac{1}{\sqrt{{\mu}_{0}{\u03f5}_{0}}}$$ | c19 |